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Question #44879

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The Question

Transition metals Zirconium (\ce{Zr}, 4d) and Hafnium (\ce{Hf}, 5d) crystallize in hexagonal close-packed (HCP) lattices at standard conditions. The shortest interatomic distance in the \ce{Zr} lattice is $320 \text{ pm}$ . A materials chemist synthesizes interstitial carbides by introducing Carbon atoms ( $r_C = 77 \text{ pm}$ ) into the octahedral voids of both metals. Based on periodic trends and geometric packing constraints, which of the following accurately describes the theoretical octahedral void radii ( $r_{void}$ ) and the mechanical nature of the resulting \ce{HfC} compared to classical expectations?

Options

A.
$r_{void}(\ce{Zr}) \approx 66.2 \text{ pm}$ and $r_{void}(\ce{Hf}) \approx 65.8 \text{ pm}$ . Carbon insertion causes severe lattice strain in \ce{HfC}, leading to unexpected brittleness, as the 5d lattice fails to provide the larger voids predicted by its higher principal quantum number.
B.
$r_{void}(\ce{Zr}) \approx 66.2 \text{ pm}$ and $r_{void}(\ce{Hf}) \approx 74.5 \text{ pm}$ . The larger 5d orbitals of \ce{Hf} expand the octahedral voids, allowing Carbon to fit with minimal strain, making \ce{HfC} significantly more malleable than \ce{ZrC}.
C.
$r_{void}(\ce{Zr}) \approx 36.0 \text{ pm}$ and $r_{void}(\ce{Hf}) \approx 35.8 \text{ pm}$ . Carbon is forced into the tetrahedral voids of both metals, but the Lanthanoid contraction in \ce{Hf} relieves internal lattice strain, increasing ductility.
D.
$r_{void}(\ce{Zr}) \approx 66.2 \text{ pm}$ and $r_{void}(\ce{Hf}) \approx 52.4 \text{ pm}$ . The Lanthanoid contraction drastically reduces the atomic radius of \ce{Hf} compared to \ce{Zr}, making interstitial carbide formation thermodynamically impossible for \ce{Hf} under standard conditions.

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How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Interstital Compound Formation|Lanthanoid Contraction cause

Why This Question Is Tricky

Common Mistakes

The core trap tests the student's ability to link Solid State Chemistry (void calculations) with Periodic Properties (Lanthanoid Contraction).

Students often assume that moving down a group (from 4d to 5d) increases the atomic radius, which would theoretically increase the size of interstitial voids and reduce lattice strain when accommodating small non-metals like Carbon.

The 'Hidden Variable' is the Lanthanoid Contraction, which causes a 'radius collapse'.

Because the 5d elements are nearly identical in size to their 4d analogs, their interstitial voids do not expand.

Forcing an interstitial atom into these constrained voids causes significant lattice distortion, explaining the unexpected mechanical brittleness of late/heavy transition metal interstitial compounds.

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Question Information

Subject	Chemistry
Chapter	D and F Block Elements - Class 12
Concept	Interstital Compound Formation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free